 The Book of Lemmas Proposition 15 - Maple Help

 Main Concept Let $\mathrm{AB}$ be the diameter of a circle, $\mathrm{AC}$ a side of an inscribed regular pentagon, $D$ the middle point of the arc $\mathrm{AC}$. Join $\mathrm{CD}$ and produce it to meet $\mathrm{BA}$ produced in $E$; join $\mathrm{AC}$, $\mathrm{DB}$ meeting in $F$, and draw $\mathrm{FM}$ perpendicular to $\mathrm{AB}$. Then, $\mathrm{EM}$ = (  ). At the beginning of the proof, you might need to note this fact:   The inscribed angle theorem states that an angle q inscribed in a circle is half of the central angle 2q, that subtends the same arc on the circle.   A consequence of this is: For four consecutive points  on a circle, .
  Proof: Let $O$ be the center of the circle and join $\mathrm{DA}$, $\mathrm{DM}$, $\mathrm{DO}$, $\mathrm{CB}$. Now:                       ,    and,         , hence,                  Further, the triangles $\mathrm{FCB}$, $\mathrm{FMB}$, are equal in all respects. Therefore, in the triangles $\mathrm{DCB}$, $\mathrm{DMB}$, the sides $\mathrm{CB}$, $\mathrm{MB}$, being equal and $\mathrm{BD}$ common, while the angles $\mathrm{CBD}$, $\mathrm{MBD}$ are equal,   However,          so that,                   , and,                        .   Therefore,                 . Now, in the triangle $\mathrm{DMO}$,                              ,                              , (because ). Therefore,               ; hence,                        .   Again:                              Therefore, in the triangles $\mathrm{EDA}$, $\mathrm{ODM}$,                                            ,                                            , and the sides $\mathrm{AD}$, $\mathrm{MD}$ are equal.   Hence, the triangles are equal in all respects and,                                                  , Therefore,                                 .